# Why does bremsstrahlung happen

1. Feb 6, 2012

### danda

1. The problem statement, all variables and given/known data

Why does it happen?

2. Relevant equations

3. The attempt at a solution

I read that Bremsstrahlung happens because a charged particle loses kinetic energy as it swings past a nucleus, therefore a photon is released to conserve energy. However, since the electric force is conservative, as an electron approaches from infinity, swings past, and leaves to infinite, the work done on it is zero.

However, this leads me to a contradiction because writing conservation of energy:

E(i) = E(f) --> E(i)=E(f) + E(photon)

where energy isn't conserved.

This seems to lead to the explanation that the electron lost energy because it emitted a photon, but this still does not answer the original question of why the photon was emitted in the first place. I know that it has something to do with charged particles being accelerated, but as long as the work done of them is zero their total energy shouldn't change so I am confused as to where the photon comes into play.

2. Feb 7, 2012

Consider X ray tubes. There is a conversion of potential energy to kinetic energy as electrons accelerate towards the anode.On reaching the anode photons are released due to the electrons losing kinetic energy by collisions.

3. Feb 7, 2012

### danda

But why? Why not just transfer their energy to the things they collide with?

And I thought it had something to do with electrons accelerating, not necessarily colliding.

4. Feb 7, 2012

There will be a considerable transfer of energy of energy to the target atoms one result of this being that the target gets hot.I think a typical X-ray tube is about 98 percent inefficient.

5. Feb 7, 2012

### danda

What is it about accelerating charges that releases the x-rays?

6. Feb 7, 2012

### Mindscrape

You're right, the brehmstralung is from decelerating electrons. In fact, apparently that's exactly what it means in German, "breaking radiation." In the most naiive sense, brehmsstrahlung is dipole radiation.

What is your argument for the work being 0? I think one fact that you're forgetting is that these electrons are moving fast, these are relativistic speeds. Once the electron slows down it's energy loss is not simply .5*m*dv^2. The electron's relativistic mass has changed.

7. Feb 7, 2012

### danda

When I say accelerated, I mean that it's direction is changed, not that it is necessarily slowed down.

And my thoughts come from the idea that the work done by an electric field is path independent. As the electron or charged particle comes from infinity, curves, and leaves to infinity, the work done on it by the atom is zero, as far as I know. My logic behind this is that on its way in through an arbitrary distance d, the work done on it is positive, while after it leaves the work done on it is negative, all the way out to that arbitrary distance d. Therefore, the net displacement is zero, and the product FΔd is zero, so the work would be zero. If this is true (which, if a photon is emitted, it seems like it can't be) the system would start with an energy E and finish with the same energy E plus the energy of the photon.

Thinking about momentum, it would seem that the atom would be also be accelerated as the electron passes by, which would keep that conserved.

8. Feb 8, 2012